Answer:
the pressure reading when connected a pressure gauge is 543.44 kPa
Explanation:
Given data
tank volume (V) = 400 L i.e 0.4 m³
temperature (T) = 25°C i.e. 25°C + 273 = 298 K
air mass (m) = 3 kg
atmospheric pressure = 98 kPa
To find out
pressure reading
Solution
we have find out pressure reading by gauge pressure
i.e. gauge pressure = absolute pressure - atmospheric pressure
first we find absolute pressure (p) by the ideal gas condition
i.e pV = mRT
p = mRT / V
p = ( 3 × 0.287 × 298 ) / 0.4
p = 641.44 kPa
so
gauge pressure = absolute pressure - atmospheric pressure
gauge pressure = 641.44 - 98
gauge pressure = 543.44 kPa
Answer:
probability P = 0.32
Explanation:
this is incomplete question
i found complete A manufactures makes integrated circuits that each have a resistance layer with a target thickness of 200 units. A circuit won't work well if this thickness varies too much from the target value. These thickness measurements are approximately normally distributed with a mean of 200 units and a standard deviation of 12 units. A random sample of 17 measurements is selected for a quality inspection. We can assume that the measurements in the sample are independent. What is the probability that the mean thickness in these 16 measurements x is farther than 3 units away from the target value?
solution
we know that Standard error is expess as
Standard error = 
Standard error =
Standard error = 3
so here we get Z value for 3 units away are from mean are
mean = -1 and + 1
so here
probability P will be
probability P = P( z < -1 or z > 1)
probability P = 0.1587 + 0.1587
probability P = 0.3174
probability P = 0.32
Answer:
The original length of the specimen 
Explanation:
Original diameter
= 30 mm
Final diameter
= 30.04 mm
Change in diameter Δd = 0.04 mm
Final length
= 105.20 mm
Elastic modulus E = 65.5 G pa = 65.5 ×
M pa
Shear modulus G = 25.4 G pa = 25.4 ×
M pa
We know that the relation between the shear modulus & elastic modulus is given by



This is the value of possion's ratio.
We know that the possion's ratio is given by


0.00476

Final length
= 105.2 m
Original length


This is the original length of the specimen.